Sobre una pista circular de atletismo de 30m de diámetro, un atleta realiza su recorrido con una velocidad tangencial de 0.5 ft/
2 answers:
Answer:
a = 0.00154 m/s²
Explanation:
On a circular running track 30m in diameter, an athlete travels with a tangential velocity of 0.5 ft / s. Determine his centripetal acceleration.
The diameter of running track is 30 m
Its radius is 15 m
The tangential velocity of the athlete is, v = 0.5 ft/s = 0.1524 m/s
We need to find the centripetal acceleration of the athlete. The centripetal acceleration is given by :
So, the centripetal acceleration of the athlete is 0.00154 m/s².
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Answer:
6.0 m/s
Explanation:
According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.
Therefore, we can write:
or
where:
m is the mass of the athlete
u is the initial speed of the athlete (at the bottom)
0 is the initial potential energy of the athlete (at the bottom)
v = 0.80 m/s is the final speed of the athlete (at the top)
is the acceleration due to gravity
h = 1.80 m is the final height of the athlete (at the top)
Solving the equation for u, we find the initial speed at which the athlete must jump: